Capacitance Help: Calculating Earth-Cloud System µC

[lostfan176]

Homework Statement

In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 35,000,000 V. The bottoms of the thunderclouds are typically 1500 m above the Earth, and can have an area of 90 km^2. For the purposes of this problem, model the Earth-cloud system as a huge capacitor. Calculate the capacitance of the Earth-cloud system in µC.

Homework Equations

C = q/V and C = K eo A/d

k= 9e9
eo = 8.85e-12

The Attempt at a Solution

9e9 * 8.85e-12 8 (90 000 000 m^2 / 1500) = 4779 C = 4779000000 µC
but this is not right

 

[foxjwill]

Why'd you include Coulomb's constant? The equation should be:

$$C=\epsilon_0 \frac{A}{d}$$

 

[lostfan176]

thanks. i don't know why put my teacher has it like that on his equation sheetCalculate the charge stored in the "capacitor" = 18.585

Calculate the energy stored in the "capacitor" =

should be 35000000 * 18.585 right?

 

[foxjwill]

Point that out to him. It's probably a typo. If it weren't, the equation would simplify to

$$\frac{A}{4\pi d}$$

because

$$k = \frac{1}{4\pi\epsilon_0}$$

 

[lostfan176]

i will.

to find energy you do w =Vq right?

 

[foxjwill]

Yup. I find it really helpful to check equations by making sure the units on both sides agree.

 

[lostfan176]

so the energy of the capacitor should be (35000000 * 18.585) 650475000 J??

 

[ylexot]

For those that are still interested in this post, there seems to be some confusion about the k in the C=κε0A/d equation. This isn't the "Coulomb constant" k = 1/(4π ε0), but rather kappa. Here κ is the dielectric coefficient of the material between the conducting plates. In our cloud example, we're generally talking about the dielectric constant of air.

 

[lolcat]

Bump!

 

[lolcat]

EDIT: Will make a new thread. Realized the question OP posted was different